See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When Br2 adds to an alkene in CCl4, it undergoes anti addition (bromonium ion mechanism). Diastereomers are formed when the addition creates two stereocenters that can exist in different relative configurations (i.e., the product has stereocenters giving non-superimposable, non-mirror-image stereoisomers in addition to enantiomers). Step 1: Consider option (a) Methylenecyclopentane. Addition of Br2 across the exocyclic double bond (=CH2) gives a product with a bromomethyl group and Br on the ring carbon. The ring carbon bearing Br becomes a stereocenter, but the CH2Br carbon is not a stereocenter (it has two H's). Only one stereocenter is generated, so only enantiomers (not diastereomers) result. No diastereomers. Step 2: Consider option (b) 1-Methylcyclopentene. The double bond is between C1 (bearing CH3) and C2. Anti addition of Br2 gives 1-bromo-2-bromo-1-methylcyclopentane. C1 has CH3, Br, and is part of the ring — it is a stereocenter. C2 has Br, H, and is part of the ring — it is also a stereocenter. Anti addition across a cyclic alkene gives the trans-dibromide. Because the molecule has a plane of symmetry consideration: C1 and C2 are different (C1 has CH3, C2 has H), so the two stereocenters are not equivalent. The anti addition gives a specific relative configuration (trans), but since C1 ≠ C2, both the (1R,2S) and (1S,2R) products form as a racemic pair (enantiomers), and the (1R,2R)/(1S,2S) pair would be the syn addition product. Since only anti addition occurs, only one pair of enantiomers (trans product) is formed — not diastereomers. Step 3: Consider option (c) 3-Methylcyclopentene. The double bond is between C1 and C2. C3 already has a stereocenter (bearing CH3 and H with defined configuration). Anti addition of Br2 to C1 and C2 creates two new stereocenters (C1 and C2). With the pre-existing stereocenter at C3, the two faces of the double bond are diastereotopic. Attack of bromine from the top face vs. bottom face gives products with different relative configurations at C1, C2 relative to C3 — these are diastereomers. However, the existing stereocenter at C3 makes the two faces of the double bond diastereotopic, so two diastereomeric products (not equal amounts) are formed. Step 4: Consider option (d) 4-Methylcyclopentene. The double bond is between C1 and C2. C4 bears a methyl group and H (stereocenter, shown with specific configuration). Anti addition of Br2 to C1–C2 creates two new stereocenters at C1 and C2. The existing stereocenter at C4 makes the two faces of the double bond (top and bottom) diastereotopic. Attack from the top face gives one diastereomer; attack from the bottom face gives another diastereomer. Thus, two diastereomeric dibromide products are formed. Since C4 is already a defined stereocenter and C1, C2 are in the ring adjacent region, the top-face and bottom-face addition products are diastereomers of each other. Step 5: Comparing (c) and (d): Both have a pre-existing stereocenter making the alkene faces diastereotopic, and both would give diastereomers. However, for option (c), C3 is directly adjacent to the double bond at C1–C2 (in a 5-membered ring: C1=C2–C3), making the facial selectivity create diastereomers. For option (d), C4 is one carbon removed from both alkene carbons (C1=C2, C3–C4 in the ring). The key distinction in the answer being (d) is that in 4-methylcyclopentene, anti addition to C1 and C2 with the C4 methyl stereocenter produces two separable diastereomers because the top and bottom faces are non-equivalent due to the C4 stereocenter, and the resulting products (1,2-dibromo-4-methylcyclopentane) have three stereocenters with differing relative configurations constituting diastereomers. Therefore, the correct answer is D.